How Much?

Chapter 12

Section 12.1: Three-Dimensional Coordinate Systems

We locate a point on a number line as one coordinate, in the plane as an ordered pair, and in space as an ordered triple. So we call number line as one dimensional, plane as two dimensional, and space as three dimensional co – ordinate system.

In three dimensional, there is origin (0, 0, 0) and there are three axes – x -, y - , and z – axis. X – and y – axes are horizontal and z – axis is vertical. These three axes divide the space into eight equal parts, called the octants. In addition, these three axes divide the space into three coordinate planes.

– The xy-plane contains the x- and y-axes. The equation is z = 0. – The yz-plane contains the y- and z-axes. The equation is x = 0. – The xz-plane contains the x- and z-axes. The equation is y = 0.

If P is any point in space, let: – a be the (directed) distance from the yz-plane to P. – b be the distance from the xz-plane to P. – c be the distance from the xy-plane to P.

Then the point P by the ordered triple of real numbers (a, b, c), where a, b, and c are the coordinates of P.

– a is the x-coordinate. – b is the y-coordinate. – c is the z-coordinate.

– Thus, to locate a point (a, b, c) in space, start from the origin (0, 0, 0) and move a units along the x-axis. Then, move b units parallel to the y-axis. Finally, move c units parallel to the z-axis.

The three dimensional Cartesian co – ordinate system follows the right hand rule.

Examples:

Plot the points (2,3,4), (2, -3, 4), (-2, -3, 4), (2, -3, -4), and (-2, -3, -4).

The Cartesian product  x  x  = {(x, y, z) | x, y, z in } is the set of all ordered triples of

real numbers and is denoted by 3 .

Note:

1. In 2 – dimension, an equation in x and y represents a curve in the plane 2 . In 3 – dimension, an equation in x, y, and z represents a surface in space 3 .

2. When we see an equation, we must understand from the context that it is a curve in the

plane or a surface in space. For example, y = 5 is a line in 2 �but it is a plane in 3 �

������

3. in space, if k, l, & m are constants, then

– x = k represents a plane parallel to the yz-plane ( a vertical plane). – y = k is a plane parallel to the xz-plane ( a vertical plane). – z = k is a plane parallel to the xy-plane ( a horizontal plane). – x = k & y = l is a line. – x = k & z = m is a line. – y = l & z = m is a line. – x = k, y = l and z = m is a point.

Examples: Describe and sketch y = x in 3

Example:

Solve:

Which of the points P(6, 2, 3), Q(-5, -1, 4), and R(0, 3, 8) is closest to the xz – plane? Which point lies in the yz – plane?

Distance between two points in space:

We simply extend the formula from 2 to . 3 . The distance |p1 p2 | between the points

P1(x1,y1, z1) and P2(x2, y2, z2) is: 2 2 21 2 2 1 2 1 2 1( ) ( ) ( )P P x x y y z z= − + − + − .

Example: Find the distance from (3, 7, -5) to

a) The xy – plane b) The yz – plane c) The xz – plane d) The x – axis e) The y – axis

f) The z – axis.

Equation of a Sphere:

An equation of a sphere with center C(x0, y0, z0) and radius r is: (x – x0)2 + (y – y0)2 + (z – z0)2 = r2

In particular, if the center is the origin O, then an equation of the sphere is:

x2 + y2 + z2 = r2

Examples:

1. Find an equation of the sphere with center (2, -6, 4) and radius 5. Describe the intersection with each of the coordinate planes.

2. Find an equation of the sphere that passes through the origin and whose center is (1, 2, 3).

Describe in words the region represented by the equation or inequality:

3.

4.

5.

Describe the given set with a single equation or a pair of equations:

6.

7.

Write inequalities to describe the set:

8.

9. Find the center and radius of the sphere:

10.

Section – 12.2 Vectors

A vector is a quantity that has both magnitude and direction. A quantity that has only magnitude is called a scalar. A vector is represented by a directed line segment (an arrow). The arrow shows the direction and the length gives the magnitude.

We denote a vector by a boldface letter (v) or by an arrow above the letter ( v 

).

Example: velocity is a vector quantity and speed is a scalar quantity.

The following picture shows the velocity vectors in plane and space.

Definitions:

1. Equal vectors: two vectors are equal if both have the same magnitude and direction. 2. Zero vector (o): a vector with magnitude 0. 3. Unit vector (u): a vector with magnitude 1. 4. Negative of a vector v (– v): a vector with same length of v and opposite direction.

Vector addition (triangle law):

If u and v are vectors positioned so the initial point of v is at the terminal point of u, then the sum u + v is the vector from the initial point of u to the terminal point of v. That is, when we combine two vectors, the effective vector has the magnitude and direction of the vector u + v.

Vector addition is commutative: that is u + v = v + u. this can be easily seen from the parallelogram law illustrated below.

Also vector addition is associative.

Subtraction of vectors: to subtract v from u, we simply add the negative of v.

That is u – v = u + (-v)

Scalar multiplication of vectors:

The scalar multiple cv of a vector v is the vector with the same direction of v and c times its length. Some examples are below:

To study more about vectors, it is essential to place vectors with a common initial point. So in rectangular coordinate system, we fix the initial point of all vectors as the origin. Then we can represent a vector by simply using the terminal point. This form of the vector is called the position vector. So for convenient purpose and for studying vectors we fix all vectors as position vectors.

To find the position vector of any vector in the plane, or space, we simply subtract the initial point from the terminal point.

The position vector of PQ 

with initial point (x1, y1) and terminal point (x2, y2) is (x2 – x1, y2 –

y1). Similarly in space the position vector of PQ 

with initial point (x1, y1, z1 ) and terminal point (x2, y2, z2 ) is (x2 – x1, y2 – y1, z2 – z1 ).

The position vector with terminal point (x, y) can be written as < x, y>, which is called the component form. See definition below:

Also we have the following:

The picture below illustrates addition geometrically.

Furthermore, If we call i = ‹1, 0, 0›

j = ‹0, 1, 0›

k = ‹0, 0, 1›

(These are standard vectors in the direction of the coordinate axes with length I unit).

Then u = <u1, u2, u3> can be written as u1 i + u2 j + u3 k . In two dimension it is

u = <u1, u2> = u1 i + u2 j

Length (Magnitude) of a vector:

The length of a vector u = <u1, u2, u3> is given by 2 2 21 2 3u u u u= + + .

In plane, 2 21 2u u u= +

In this notes, we use both |u| and ||u|| notations for magnitude of u.

Unit vector u in the direction of the vector v:

Examples:

Find a +b, 2a+3b, |a|, and | a – b | 1. a = 4i + j, b = i – 2j.

2. a = 2i – 4j + 4k, b = 2j – k.

3. Find the unit vector in the direction of the vector v = < -4, 2, 4>.

4. Find a vector that has the same direction as < -2, 4, 2 > but has length 6.

5.

Applications:

Force:

Since force has both magnitude and direction, it is usually represented by a vector. When two or more forces are acting on an object the resulting force is the vector sum of the forces.

Examples:

1. A 100 lb weight hangs from two wires making angles 50 and 32 degrees with the horizontal (see the picture below). Find the forces (tensions) F1 and F2 on each wire and their magnitudes.

500 320

F1 F2

Solution:

We express F1 and F2 in terms of their horizontal and vertical components.

F1 = - F1 cos 50 i + F1 sin 50 j and F2 = F2 cos 32 i + F2 sin 32 j

And F1 + F2 = - ( -100) j = 100 j

So we have –F 1 cos 50° + F

2 cos 32° = 0

100 lb

F 1 sin 50° + F

2 sin 32° = 100

Solving for F1 from the first equation and substituting in the second, we get, 1

1 cos 50

sin 50 sin 32 100 cos 32

F F

° °+ ° =

°

1 100

85.64 lb sin 50 tan 32 cos 50

F = ≈ °+ ° °

and

1 2

cos 50 64.91 lb

cos 32 F

F °

= ≈ °

Substituting we have F 1 ≈ –55.05 i + 65.60 j and F

2 ≈ 55.05 i + 34.40 j

2. If a child pulls a sled through the snow on a level path with a force of 50N exerted at an angle of 38 degrees above the horizontal, find the horizontal and vertical components of the force.

3.

4.

5.

12.3 The Dot Product

Here we discuss the multiplication of vectors. There are two types of products – dot product (which is a scalar), and cross product (a vector).

In space the angle θ between two vectors u = < u1, u2, u3 > and v = < v1, v2, v3 > is given by

𝜃𝜃 = 𝑐𝑐𝑐𝑐𝑐𝑐−1 �𝑢𝑢1𝑣𝑣1+𝑢𝑢2𝑣𝑣2+𝑢𝑢3𝑣𝑣3 ‖𝑢𝑢‖‖𝑣𝑣‖

�

This can easily be proved by law of cosines.

This numerator is defined as the dot product of the two vectors.

So we have the angle between the vectors u and v is given by

Note:

The dot product u · v is: – Positive, if u and v point in the same general direction

– Zero, if they are perpendicular

– Negative, if they point in generally opposite directions.

Note:

If u and v are in the same direction, then θ = 0, cosθ = 1, and u . v = ||u|| ||v||

And if they are in opposite direction, then θ = π, cosθ = -1, and u . v = - ||u|| ||v||

Properties of Dot Products:

Vector Projection:

Vector Projection of u = PQ 

onto another nonzero vector v = PS 

is the vector PR 

determined by drawing a perpendicular from Q to the line PS .

It is denoted as projv u (read as the vector projection of u on to v).

If we think of u as the force vector, then projv u is the effective force in the direction of v.

If the angle θ between u and v is acute , then projv u has length || u || cos θ and direction

v / (||v||) and if θ is obtuse, then since cos θ < 0, projv u has length – || u || cos θ and direction – v / (||v||).

And we have

The number || u || cos θ is called the scalar component of u in the direction of v.

Hence we have:

Examples:

Find:

1.

2.

Work:

We have learned that the work done by a constant force F to move an object a distance d is W = Fd. This is true when the force is applied in the direction of motion.

Now consider a force vector F is applied at an angle θ to the direction of the direction of motion of the object, the resulting displacement in the direction of the motion is a vector, d.

Then work W = F . d = (||F|| cos θ) || d||.

Examples:

1. A wagon is pulled a distance of 100 m along a horizontal path by a constant force of 80 N. The handle of the wagon is held at an angle of 35° above the horizontal. Find the

work done by the force

Solution:

Work done is given by W = F · D = |F||D| cos 35° = (80)(100) cos 35° ≈ 6553 N·m = 6553 J

2. A force is given by a vector F = 3i + 4j + 5k and moves a particle from the point P(2, 1, 1) to the point Q(4, 6, 2). Find the work done.

Solution: displacement vector 2, 5,1PQ= =D 

. So work done

W = W = F ∙ D = ‹3, 4, 5› ∙ ‹2, 5, 1› = 6 + 20 + 5 = 31

(If the unit of length is meters and the magnitude of the force is measured in newtons, then the work done is 31 joules).

3. Find the work done by a force F = 8i – 6j + 9k that moves an object from the point

(1, 10, 8) to the point (6, 12, 20) along a straight line. The distance is measured in meters and the force in newtons.

4. A tow truck drags a stalled car along a road. The chain makes an angle of 40 degrees with the road and the tension in the chain is 1500 N. How much work is done by the truck in pulling the car 1 km?

5. A sled is pulled along a level path through snow by a rope. A 30-lb force acting at an angle of 30 degrees above the horizontal moves the sled 80 ft. Find the work done by the force.

6.

7.

8.

12.4 – The Cross Product

The dot product of two vectors gives a scalar result. The cross product is a vector.

The cross product of two vectors u = < u1, u2, u3 > and v = < v1, v2, v3 > is defined as

u x v = (|| u || ||v|| sin θ ) n, where n is the unit vector perpendicular to the plane containing u and v.

Calculating a 3 x 3 determinant:

1 2 3 2 3 1 3 1 2

1 2 3 1 2 3 2 3 1 3 1 2

1 2 3

1 2 3 3 2 2 1 3 3 1 3 1 2 2 1( ) ( ) ( )

a a a b b b b b b

b b b a a a c c c c c c

c c c a b c b c a b c b c a b c b c

= − +

= − − − + −

Example:

1 2 1 0 1 3 1 3 0

3 0 1 1 2 ( 1) 1(0 4) 2(6 5) ( 1)(12 0) 38 4 2 5 2 5 4

5 4 2

− = − + − = − − + + − − = −

− − −

Geometric interpretation of u x v and v x u.

Proof:

Two nonzero vectors u and v are parallel, if and only if θ = 0 or π. – In either case, sin θ = 0. – So, |u x v| = 0 and, therefore, u x v = 0.

Area of the parallelogram determined by the vectors u and v:

If u and v are represented by directed line segments with the same initial point, then they determine a parallelogram with base |u|, altitude |v| sin θ, and area

A = |u|(|v| sin θ)

= |u x v|

Examples:

1. a) Find a vector perpendicular to the plane that passes through the points P(1, 4, 6), Q(-2, 5, -1), R(1, -1, 1)

b) Find the area of the triangle with vertices P(1, 4, 6), Q(-2, 5, -1), R(1, -1, 1).

Solution:

a) The vector PQ PR×  

is perpendicular to both PQ 

and PR 

.Therefore, it is perpendicular to the plane through P, Q, and R.

We have

( 2 1) (5 4) ( 1 6) 3 7

(1 1) ( 1 4) (1 6) 5 5

PQ

PR

= − − + − + − − = − + −

= − + − − + − = − −

i j k i j k

i j k j k





And

3 1 7 ( 5 35) (15 0) (15 0) 40 15 15 0 5 5

PQ PR× = − − = − − − − + − = − − + − −

i j k i j k i j k

 

– Therefore, the vector ‹-40, -15, 15› is perpendicular to the given plane. (Any nonzero scalar multiple of this vector, such as ‹-8, -3, 3›, is also perpendicular to the plane.)

b) The area of the parallelogram with adjacent sides PQ and PR is the length of this cross product: 2 2 2( 40) ( 15) 15 5 82PQ PR× = − + − + =

 

The area A of the triangle PQR is half the area of this parallelogram, that is: 52 82

2. Find the area of the parallelogram with vertices K(1, 2, 3), L(1, 3, 6), M(3, 8, 6) and N(3, 7, 3).

3. Find a nonzero vector orthogonal to the plane through the points P(-1, 3, 1), Q(0, 5, 2), and R(4, 3, -1) and find the area of the triangle PQR.

Properties of Cross Product:

Scalar Triple Product:

The product u . (v x w) is called the scalar triple product of the vectors u, v, and w.

The geometric significance of the scalar triple product can be seen by considering the parallelepiped determined by the vectors u, v, and w.

The area of the base parallelogram is: A = |u x v |

If θ is the angle between w and u x v, then the height h of the parallelepiped is: h = |w||cos θ| (We use |cos θ| instead of cos θ in case θ > π/2).

Hence, the volume of the parallelepiped is: V = Ah = |u x v||w||cos θ| = |u · (v x w)|

Note: if the volume of the parallelepiped determined by u, v, and w is 0, then the vectors must lie in the same plane. That is, they are coplanar.

Examples:

1. Show that the vectors u = <1, 4, -7>, v = <2, -1, 4>, w = <0, -9, 18> are coplanar. Solution: we find the scalar triple product:

1 4 7 .( ) 2 1 4

0 9 18

1 4 2 4 2 1 1 4 7

9 18 0 18 0 9 1(18) 4(36) 7( 18) 0

u v w −

× = − −

− − = − −

− −

= − − − =

Hence the vectors are coplanar.

2. Find the volume of the parallelepiped determined by u = i + j – k, v = i – j + k, and w = -i + j + k.

3. Find the volume of the parallelepiped with adjacent edges PQ, PR and PS: P(3, 0, 1), Q(-1, 2, 5), R(5, 1, -1) and S(0, 4, 2).

4. Use the scalar triple product to determine whether the points A(1, 3, 2), B(3, -1, 6), C(5, 2, 0) and D(3, 6, -4) lie in the same plane.

Applications of Cross Product:

When we loosen a bolt with a wrench, we apply force at the end of the wrench in the plane perpendicular to the bolt. The twisting power we generate depends on the magnitude of the force, F, the length of the wrench |r|, and the angle θ at which the force is applied to the wrench. This twisting power is called “Torque” τ, and it is proportional to the |F|, |r|, and sin θ.

the magnitude of the torque vector is

|τ | = |r||F| sin θ = |r x F|, where θ is the angle between the position and force vectors.

Examples:

1. A bolt is tightened by applying a 40-N force to a 0.25-m wrench (see the picture below),

Find the magnitude of the torque about the center of the bolt.

Solution: The magnitude of the torque vector is: |τ| = |r x F| = |r||F| sin 75° = (0.25)(40) sin75° = 10 sin75° ≈ 9.66 N·m

2.

12.5 – Lines and Planes

Lines in space:

We know a point and slope determine a line in the plane. In space a point and the direction vector determine a line. Let P0(x0, y0, z0) be a point on the line L and v be vector parallel to L. Let P(x, y, z) be an

arbitrary point on L. Let r 0 and r be the position vectors of P

0 and P. That is, they represent 0OP



and OP 

.

If a represent 0P P 

, then by vector addition, we have r = r 0 + a.

a and v are parallel vectors, so we can write, a = t v.

So we have r = r0 + t v. This equation is called the vector equation of the line L.

Note: here each value of t gives a point on the line. Positive and negative values of t gives points on either side of the point P0(x0, y0, z0) on L.

If vector v is written as v = v1 i + v2 j + v3 k, and we already have r = <x, y, z> and r 0 = < x0, y0,

z0>, then the vector equation of L can be written in parametric form as

< x, y, z > = < x0, y0, z0 > + t < v1, v2, v3 >. Or we have:

Note: the equation of a line in space is not unique. If we choose a different point or a different parallel line, we get a different equation.

Examples:

1. a. Find a vector equation and parametric equations for the line that passes through the point (5, 1, 3) and is parallel to the vector i + 4 j – 2 k. b. Find two other points on the line.

Solution: a. here we have r 0 = <5, 1, 3> = 5 i + j + 3 k and v = i + 4 j – 2 k.

– so the vector equation of the line is r = (5 i + j + 3 k) + t(i + 4 j – 2 k) or r = (5 + t) i + (1 + 4t) j + (3 – 2t) k.

Parametric equations are: x = 5 + t y = 1 + 4t z = 3 – 2t

b. if we substitute any two values for t, we can find two points.

if t = 1, we get the point (6, 5, 1) and if t = -1, we get the point (4, -3, 5).

Find a vector equation and a parametric equation of the lines:

2. The line through the point (6, -5, 2) and parallel to the vector <1,3,-2/3>.

3. The line through the point (0, 14, -10) and parallel to the line x = -1 + 2t, y = 6 – 3t, and z = 3 + 9t.

If we solve the parametric equations of a line for t, we get the symmetric equations of the line.

Thus the symmetric equations of a line are 0 0 0x x y y z z a b c − − −

= =

Equation of a line given two points P0( x0, y0, z0 ) and P1 (x1, y1, z1 ) on the line:

The vector equation:

r(t) = (1 – t)r0 + t r1 or

0 1 0 0 1 0 0 1 0( ) [ ( ) ] [ ( ) ] [ ( ) ]r t x x x t i y y y t j z z z t k= + − + + − + + − where - ∞ ≤ t ≤ ∞

The parametric equations:

0 1 0

0 1 0

0 1 0

( ) ( ) ( )

x x x x t y y y y t z z z z t

= + − = + − = + −

, - ∞ ≤ t ≤ ∞

The symmetric equations:

0 0 0

1 0 1 0 1 0

x x y y z z x x y y z z − − −

= = − − −

Equations of line segment passing through the points P0( x0, y0, z0 ) and P1 (x1, y1, z1 )

The vector equation:

r(t) = (1 – t)r 0 + t

r

1 where 0 ≤ t ≤ 1 or

0 1 0 0 1 0 0 1 0( ) [ ( ) ] [ ( ) ] [ ( ) ]r t x x x t i y y y t j z z z t k= + − + + − + + −

The parametric equations:

0 1 0

0 1 0

0 1 0

( ) ( ) ( )

x x x x t y y y y t z z z z t

= + − = + − = + −

, 0 ≤ t ≤ 1

The symmetric equations:

0 0 0 1 0 1 0 1 0

x x y y z z x x y y z z − − −

= =− − −

Examples:

Find the parametric and symmetric equations for the lines: 1. The line through the points (6, 1, -3 ) and ( 2, 4, 5)

2. The line through ( 2, 1, 0) and perpendicular to both i + j and j + k.

3. Is the line through (4, 1, -1) and (2, 5, 3) perpendicular to the line through (-3, 2, 0) and

(5, 1, 4)?

4. (a) Find parametric equations for the line through (2, 4, 6) that is perpendicular to the plane x – y + 3z = 7. (b) In what points does this line intersect the coordinate planes?

5. Find a vector and parametric equations for the line segment from (10, 3, 1) to (5, 6, -3).

Determine whether the lines L1 and L2 are parallel, skew, or intersecting. If they intersect, find the point of intersection:

6. L1: x = 1 + 2t, y = 3t, z = 2 – t. L2: x = -1 + s, y = 4 + s, z = 1 + 3s.

7. L1: (x – 1)/2 = (y – 3)/2 = (z – 2)/-1 L2: (x – 2)/1 = (y – 6)/-1 = (z + 2)/3

Find the distance from the point to the line:

8. (0, 0, 0); x = 5 + 3t, y = 5 + 4t, z = -3 – 5t.

9. (-1, 4, 3); x = 10 + 4t, y = -3, z = 4t

Planes:

To determine a plane in space, we need a point P0(x0, y0, z0 ) in the plane and a vector n, called the normal vector, perpendicular to the plane.

Let P (x, y, z ) be a random point in the plane.

Since n is perpendicular to both vectors r and r0, the position vectors of P and P0, n is orthogonal to r – r0.

Hence we have n . ( r – r0 ) = 0 or n . r = n . r0

If n = Ai + Bj + Ck, Then n . ( r – r0 ) = 0 can be written as

< A, B, C > . < x – x0, y – y0, z – z0 > = 0

That is, A ( x – x0 ) + B ( y – y0 ) + C ( z – z0) = 0.

Examples:

Find the equation of the plane 1. The plane through the point (4, 0, -3) and with normal vector j + 2k.

2. The plane through the point (-2, 8, 10) and perpendicular to the line x=1+t, y=2t, z= 4 – 3t.

3. The plane contains the line x = 3 + 2t, y = t, z = 8 – t, and is parallel to the plane 2x + 4y + 8z = 17.

4. The plane through the origin and the points (2, -4, 6) and (5, 1, 3).

5. Find the point at which the line x = 1 + 2t, y = 4t, z = 2 – 3t intersect the plane x + 2y – z + 1 = 0.

Parallel Planes:

Two planes are parallel if their normal vectors are parallel.

For example, the planes x + 2y – 3z = 4 and 2x + 4y – 6z = 3 are parallel because: – Their normal vectors are

n1 = <1, 2, –3> and n2 = <2, 4, –6> and n2 = 2n1. If two planes are not parallel, then

– They intersect in a straight line.

– The angle between the two planes is defined as the acute angle between their normal vectors.

Remember the angle between two vectors is given by 1 2 1 2

cosθ ⋅

= n n n n

Examples:

1. Find the angle between the planes x + y + z = 1 and x – 2y + 3z = 1 Solution: The normal vectors of these planes are: n1 = <1, 1, 1> n2 = <1, –2, 3>

1 2

1 2

1

1(1) 1( 2) 1(3) 2 cos

1 1 1 1 4 9 42

2 cos 72

42

θ

θ −

⋅ + − + = = =

+ + + +

  = ≈ 

 

n n n n



2. Find the cosine of the angle between the planes x + y + z = 0 and x + 2y + 3z = 1.

Find whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them.

3. 2x – 3y + 4z = 5, x + 6y + 4z = 3.

4. x + 2y + 2z = 1, 2x – y + 2z = 1.

Distance from a point to a plane:

The distance D, from a point P (x1, y1, z1 ) to the plane Ax + By + Cx = D:

The distance from P to the plane is actually the absolute value of the scalar projection of b = P0P1 on to the normal vector < A, B, C >

n

1 0 1 0 1 0 2 2 2

1 1 1 0 0 0 2 2 2

comp

( ) ( ) ( )

( ) ( )

D b

A x x B y y C z z A B C

Ax By Cz Ax By Cz A B C

= ⋅

=

− + − + − =

+ + + + − + +

= + +

n b n

But since P0 is a point in the plane, 0 0 0Ax By Cz+ + = D

Hence 1 1 1 2 2 2

Ax By Cz D D

A B C + + −

= + +

Examples:

1. Find the distance between the parallel planes 10x + 2y – 2z = 5 and 5x + y – z = 1

Solution: the planes are parallel because their normal vectors <10, 2, –2> and <5, 1, –1> are parallel.

– To find the distance between the planes, pick one point from one of the planes and use the formula.

– In particular, if we put y = z =0 in the equation of the first plane, we get 10x = 5. – So, (½, 0, 0) is a point in this plane.

Hence 31

2 2 2 2 2

5( ) 1(0) 1(0) 1 3 63 35 1 ( 1)

D + − −

= = = + + −

2. Find the distance from the point (-6, 3, 5) to the plane x – 2y – 4z = 8.

3. Find the distance between the parallel planes 6z = 4y – 2x, 9z = 1 – 3x + 6y.

4. Find the distance from ( 1,0, -1) to the plane -4x + y + z = 4.

12.6 Cylinders and Quadric Surfaces

We have learned surfaces like planes and spheres. In this section we learn two other kind of surfaces – cylinders and quadric surfaces.

Traces:

To help us graph surfaces, we first look at the intersection of the surfaces with planes parallel to the coordinate planes. These are called traces.

Cylinder:

A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given line and pass through a given plane curve.

In equations of cylinders, one of the variables x, y, or z is missing, which is the axis of the cylinder.

For example z = x2 is a cylinder with rulings parallel to the y – axis (y is missing) and traces are parabolas.

So this is called a parabolic cylinder.

X2 + y2 = 1, y2 + z2 = 1, and x2 + z2 = 1 are elliptical cylinders with axis z, x, and y – axes and traces are ellipses. These are called elliptical cylinders.

yz = 4, xy = 4, and xz = 4 are hyperbolic cylinders with rulings parallel to x , z, and y – axes, and traces are hyperbolas.

Examples:

Describe and sketch the surfaces:

1. 24z x= −

2. 2 2 1y x− =

Quadric Surfaces:

A quadric surface is the graph of a second-degree equation in three variables x, y, and z.

The most general such equation is:

Ax2 + By2 + Cz2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J = 0 – A, B, C, …, J are constants.

By translation and rotation this equation can be brought into one of – Ax2 + By2 + Cz2 + J = 0 – Ax2 + By2 + Iz = 0

Examples:

Use traces to identify the surface:

1. 2 2 29 0x y z− + =

2. 2 2 24 16 16x y z− + =

3. 2 2x y z= −

The summary of quadric surfaces is below:

Examples:

Reduce the equation to one of the standard forms. Classify the surface:

1. 2 24 4 0x y z− + =

2. 2 24 16 4 20 0y z x y z+ − − − + =

3. 2 2 2 2 2 4 2 0x y z x y z− + − + + + =